Infinite sequences of PPT’s, generated by relatively simple formulae, can be found in a binary vine-like structure which sits ‘inside’ the ternary tree structure of PPT’s discovered by B. Berggren in 1934.

On reading an excerpt from a book ‘The Pythagorean Theorem: A 4,000-Year History (2007) by Eli Maor’ there is mention of the PPT (4601,4800,6649) which was/is stamped into the clay tablet designated Plimpton 322.

It is straightforward to calculate where this PPT sits within the ‘vine-like’ structure of all PPT’s.

(4601,4800,6649) is the first PPT (n=1) on the branch d=2(c+a) seeded by (704,903,1145) = (a,b,c).

Likewise, (704,903,1145) is the first PPT (n=1) on the branch d=2(c+a) seeded by (21,220,221) = (a,b,c).

(21,220,221) is at position n=10 on the main stem.

So, to get to (4601,4800,6649) from the root (0,1,1), take 10 steps along the main stem to (21,220,221), then take one step down the d=2(c+a) branch seeded by (21,220,221) to (704,903,1145), followed by one step down the d=2(c+a) branch seeded by (704,903,1145) to (4601,4800,6649).

Some identities

The ‘main stem’ sequence of PPT’s is just (2n+1, n(2n+1)+n, n(2n+1) +n + 1), n∈ω, where 2n+1 is any odd positive integer.

If we choose (a,b,c) = (3,4,5) as a seed PPT, then choose d=2(c+a) we get the generated sequence of PPT’s (16n+4,n(16n+8)-3,n(16n+8)+5) = (16n+4,16n²+8n-3,16n²+8n+5),n∈ω.

So, we have from the main stem, (2n+1)² = (2n² + 2n + 1)² – (2n² + 2n)² , while we also have from (16n+4,16n²+8n-3,16n²+8n+5), n∈ω that (16n²+8n-3)² = (16n²+8n+5)² – (16n+4)².

Now, 16n²+8n-3 = 2(8n²+4n-2) + 1 and so we can replace n by 8n²+4n-2 in